Mixing
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}exp^{{−0.5...
$begingroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.
a) Find the mean and the variance of $X$.
I know that $int_0^infty f(x) dx=1$, but this says "consider the random variable $X$ with the log-normal pdf" so would I actually be integrating $int_0^infty x f(x) dx$ for the mean? The reason I ask (and I'm skeptical) is because the solution to that integral is disgusting and has $i$ in the solution and that doesn't seem to be applicable for what I'm doing. Any help is greatly appreciated.
integration statistics random-variables statistical-inference means
asked Jan 26 at 2:24
ddswsdddswsd
37929
$endgroup$
add a comment |
$begingroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.
a) Find the mean and the variance of $X$.
I know that $int_0^infty f(x) dx=1$, but this says "consider the random variable $X$ with the log-normal pdf" so would I actually be integrating $int_0^infty x f(x) dx$ for the mean? The reason I ask (and I'm skeptical) is because the solution to that integral is disgusting and has $i$ in the solution and that doesn't seem to be applicable for what I'm doing. Any help is greatly appreciated.
integration statistics random-variables statistical-inference means
asked Jan 26 at 2:24
ddswsdddswsd
37929
$endgroup$
$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
1
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57
add a comment |
$begingroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.
a) Find the mean and the variance of $X$.
I know that $int_0^infty f(x) dx=1$, but this says "consider the random variable $X$ with the log-normal pdf" so would I actually be integrating $int_0^infty x f(x) dx$ for the mean? The reason I ask (and I'm skeptical) is because the solution to that integral is disgusting and has $i$ in the solution and that doesn't seem to be applicable for what I'm doing. Any help is greatly appreciated.
integration statistics random-variables statistical-inference means
asked Jan 26 at 2:24
ddswsdddswsd
37929
$endgroup$
Consider a random variable $X$ with the log-normal pdf $f(x) ={1over sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.
a) Find the mean and the variance of $X$.
I know that $int_0^infty f(x) dx=1$, but this says "consider the random variable $X$ with the log-normal pdf" so would I actually be integrating $int_0^infty x f(x) dx$ for the mean? The reason I ask (and I'm skeptical) is because the solution to that integral is disgusting and has $i$ in the solution and that doesn't seem to be applicable for what I'm doing. Any help is greatly appreciated.
integration statistics random-variables statistical-inference means
integration statistics random-variables statistical-inference means
asked Jan 26 at 2:24
ddswsdddswsd
37929
asked Jan 26 at 2:24
ddswsdddswsd
37929
asked Jan 26 at 2:24
ddswsdddswsd
37929
asked Jan 26 at 2:24
ddswsdddswsd
37929
asked Jan 26 at 2:24
ddswsdddswsd
37929
37929
$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
1
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57
add a comment |
$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
1
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57
$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
1
1
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
hint: it is $int_0^{infty} xf(x), dx$. To evaluate this make the substitution $y=log, x$. And then use the following: $-frac 12 y^{2}+y=-frac 1 2 (y-1)^{2} +frac 1 2$. You should get the answer as $sqrt e$. [ You have to use the fact $frac 1 {sqrt {2pi}} int_{-infty}^{infty} e^{-frac 1 2 y^{2}}, dy =1$]
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
hint: it is $int_0^{infty} xf(x), dx$. To evaluate this make the substitution $y=log, x$. And then use the following: $-frac 12 y^{2}+y=-frac 1 2 (y-1)^{2} +frac 1 2$. You should get the answer as $sqrt e$. [ You have to use the fact $frac 1 {sqrt {2pi}} int_{-infty}^{infty} e^{-frac 1 2 y^{2}}, dy =1$]
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
|
show 2 more comments
$begingroup$
hint: it is $int_0^{infty} xf(x), dx$. To evaluate this make the substitution $y=log, x$. And then use the following: $-frac 12 y^{2}+y=-frac 1 2 (y-1)^{2} +frac 1 2$. You should get the answer as $sqrt e$. [ You have to use the fact $frac 1 {sqrt {2pi}} int_{-infty}^{infty} e^{-frac 1 2 y^{2}}, dy =1$]
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
|
show 2 more comments
$begingroup$
hint: it is $int_0^{infty} xf(x), dx$. To evaluate this make the substitution $y=log, x$. And then use the following: $-frac 12 y^{2}+y=-frac 1 2 (y-1)^{2} +frac 1 2$. You should get the answer as $sqrt e$. [ You have to use the fact $frac 1 {sqrt {2pi}} int_{-infty}^{infty} e^{-frac 1 2 y^{2}}, dy =1$]
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
hint: it is $int_0^{infty} xf(x), dx$. To evaluate this make the substitution $y=log, x$. And then use the following: $-frac 12 y^{2}+y=-frac 1 2 (y-1)^{2} +frac 1 2$. You should get the answer as $sqrt e$. [ You have to use the fact $frac 1 {sqrt {2pi}} int_{-infty}^{infty} e^{-frac 1 2 y^{2}}, dy =1$]
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
edited Jan 26 at 5:23
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
answered Jan 26 at 5:13
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
68.8k53169
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
|
show 2 more comments
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
Should the substitution be $y=log x$?
$endgroup$
– ddswsd
Jan 26 at 5:19
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
@ddswsd Yes, sorry for the typo.
$endgroup$
– Kavi Rama Murthy
Jan 26 at 5:24
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
If I let $y= log x$, then the integral becomes ${1over sqrt{2pi}}int_{- infty} ^{infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $sqrt{e}$.
$endgroup$
– ddswsd
Jan 26 at 14:16
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
@ddswsd The integral you have written is $sqrt e$ as far as I can see. Can you show me how you evaluated it?
$endgroup$
– Kavi Rama Murthy
Jan 26 at 23:26
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
$begingroup$
Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $sqrt{e}$. Then I changed the decimals to fractions and the answer is $sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you!
$endgroup$
– ddswsd
Jan 27 at 0:13
|
show 2 more comments
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$begingroup$
See here for some approaches for the mean: math.stackexchange.com/questions/2409702/…
$endgroup$
– jjjjjj
Jan 26 at 2:28
1
$begingroup$
Disgusting?!? :)
$endgroup$
– Aditya Dua
Jan 26 at 5:57